Really confused, pls help asap

Question

factorise each of the following expressions completely  

 

a) 6x^2y^2-5xy-4

 

b) 3z-8xyz+4x^2y^2z

 

c) 36y^2-49(x+1)^2

 

Answer 1

If you rewrite the problems and substitute x*y by t for example, you'll get a quadratic equation. 

for b)  part you have the same after factoring out z. 

c) part is a²-b² problem.

 

Do you need more information on that?

Answer 2

Hi Amy S

After putting the terms of your expression into proper order from highest exponent to lowest, you can factor by grouping

a) 6x²y² - 5xy - 4

Multiply the coefficient of the squared variables by the constant that would be (6)(4)= -24

With a focus on the middle term list the factors of -24 that multiply to -24 but combine to -5

-24 = (-8)(3) and -8 + 3 = -5

Replace the middle term with -8xy and 3xy as follows:

6x²y² - 8xy + 3xy - 4

Group into pairs

(6x²y² - 8xy) + (3xy -4)

Factor the pairs

2xy(3xy -4) + 1(3xy - 4)

Notice (3xy -4) is the common binomial in the expression above which can be further factored

(3xy - 4)(2xy + 1)

You can always use FOIL to check that your factoring is correct.

b) Put the exponents in the proper order

b) 4x²y²z - 8xyz + 3z

factor out z since it is common to all terms in the expression

z(4x²y² - 8xy + 3)

Remember for this one z stays along throughout the process

Now repeat the process used for question (a). (4)(3) = 12 what factors of 12 multiply to 12 but add to -8

(-2)(-6) = 12, so now just replace -8xy with -2xy - 6xy

z(4x²y² - 2xy - 6xy + 3)

Group into pairs

z((4x²y² - 2xy) - (6xy + 3))

Factor each pair

z(2xy(2xy - 1) - 3(2xy - 1))

(2xy - 1) is the common binomial of the expression which

z((2xy - 3)(2xy - 1))

Again you can check (a) and (b) by FOIL. Question (c) is the special case as described by Marina K above as the Difference between two perfect squares. I hope you find this useful.